Problem: $A=\left[\begin{array}{rr}2 & -1 & 8 & 15 & 1 \\6 & -12 & -13 & 0 & 5 \\0 &-5 &1 & 9 & -8 \\3 &7 &-2 & -1 & 9\end{array}\right]$ $A_{3,5}=$
Background An $m\times n$ matrix has $m$ rows and $n$ columns. $A=\left[\begin{array}{rr}A_{1,1} & \cdots & A_{1,n} \\\\\vdots \ & \ddots & \vdots \\\\A_{m,1} &\cdots &A_{m,n}\end{array}\right]$ Therefore, the entry $A_{{c},{d}}$ is located on row ${c}$ and column ${d}$. Finding $A_{3,5}$ $A_{{3},{5}}$ is located on row ${3}$ of $A$ : $\left[\begin{array}{rr}2 & -1 & 8 & 15 & 1 \\6 & -12 & -13 & 0 & 5 \\ {0} & {-5} & {1} & {9} & {-8} \\3 &7 &-2 & -1 & 9\end{array}\right]$ $A_{{3},{5}}$ is also located on column ${5}$ of $A$. $\left[\begin{array}{rr}2 & -1 & 8 & 15 & 1 \\6 & -12 & -13 & 0 & 5 \\ {0} & {-5} & {1} & {9} & {\text{-8}} \\3 &7 &-2 & -1 & 9\end{array}\right]$ Therefore, $A_{{3},{5}}={-8}$. Summary $A_{3,5}=-8$